Fermentatia alcoolica: descompunerea glucozei in alcool etilic si CO2, de aici termenul de alcoolica
calculam masa de alcool dizolvata in solutie (md)
c=md*100/ms => md=c*ms/100=15*500/100=75kg alcool etilic
scriem reactia si asezam pe reactie ce stim. Alcoolul etilic are urmatoarea structura: CH3-CH2-OH insa il voi scrie cu formula moleculara: C2H6O
Mglucoza=12*2+6+16*6=180kg/kmol
Malcool=12*2+6+16=46kg/kmol
180kg............2*46kg
C6H12O6 -> 2C2H6O + 2CO2
y=?..................75kg
y=75*180/92=146,73kg glucoza
